Compound interest

Simple interest pays you the same rupee amount every year, because it is always calculated on the original principal $P$. Compound interest is different: at the end of each period, the interest earned is added to the principal, and the next period's interest is calculated on this new, larger sum. The gap between CI and SI is zero in the first year (there has been no "interest on interest" yet) and widens every year after that.

Method 1 — successive years. Treat each year as its own simple-interest step, using the previous year's Amount as the new principal: find $SI_1$ on $P$, get $A_1 = P + SI_1$; find $SI_2$ on $A_1$ (not on $P$), get $A_2 = A_1 + SI_2$; and so on.

Method 2 — direct formula. For a principal $P$, annual rate $R\%$, compounded yearly for $n$ years:

where: $A$ is the Amount after $n$ years, $P$ is the original principal, $R$ is the rate per cent per annum, and $n$ is the number of years (up to 3 at this level). For half-yearly compounding, interest is added twice a year, so the rate per period is halved and the number of periods is doubled. With annual rate $R\%$ over $n$ years:

where: $2n$ is the number of half-year steps (up to 3 half-year steps at this level), and $R/200$ is the half-yearly rate written as a fraction of $100$. Once $A$ is known, the compound interest is simply

For the same $P$, $R$ and $T$, $CI \ge SI$ always, with equality only at $T = 1$ year. In the first year both methods charge interest on exactly the same principal $P$, so they give the same rupee amount. From the second year onward, SI keeps charging interest on the unchanging $P$, while CI charges interest on an Amount that already includes earlier interest — so CI edges ahead, and the gap between the two widens with every additional year, faster than it might look from the numbers alone.

Find the compound interest on ₹8,000 for 2 years at 10% per annum, compounded yearly, and check it against the direct formula.

Step 1. Find the Amount after Year 1 using simple interest on the original principal. $SI_1 = \dfrac{8{,}000 \times 10 \times 1}{100} = 800$, so $A_1 = 8{,}000 + 800 = 8{,}800$.

Step 2. Find the Amount after Year 2, using $A_1 = 8{,}800$ as the new principal, not the original ₹8,000. $SI_2 = \dfrac{8{,}800 \times 10 \times 1}{100} = 880$, so $A_2 = 8{,}800 + 880 = 9{,}680$.

Step 3. Find the compound interest by subtracting the original principal from the final Amount. $CI = A_2 - P = 9{,}680 - 8{,}000 = 1{,}680$.

Step 4. Check with the direct formula. $A = 8{,}000\left(1+\dfrac{10}{100}\right)^2 = 8{,}000 \times 1.1^2 = 8{,}000 \times 1.21 = 9{,}680$, matching Step 2 exactly.

The compound interest for 2 years is ₹1,680.

Competitive stretch

Extend to half-yearly compounding: find the amount and CI on ₹40,000 for $1\tfrac12$ years at 8% per annum, compounded half-yearly. The half-yearly rate is $\dfrac{8}{2} = 4\%$ and the number of half-year steps is $1.5 \times 2 = 3$. So $A = 40{,}000(1.04)^3 = 40{,}000 \times 1.124864 = 44{,}994.56$, giving $CI = 44{,}994.56 - 40{,}000 =$ ₹4,994.56.

• Using the annual rate directly for half-yearly compounding. The rate must be halved and the time doubled; skipping either step gives a badly wrong Amount.
• Using the original principal for every year in the successive-years method. Each new year's interest is calculated on the previous Amount, not on $P$ — using $P$ every time silently turns CI back into SI.
• Arithmetic slips while raising $(1+r)$ to a power. Since $n$ is small (up to 3), compute the power one multiplication at a time and cross-check with the successive-years method.
• Forgetting to subtract $P$ at the end. $A = P(1+R/100)^n$ gives the Amount, not the interest; CI is found only after subtracting $P$.

Ask the tutor: if converting an annual rate and time into the matching half-yearly rate and number of steps feels confusing, ask the AI tutor to walk through a fresh example with you.